Given an input string (s
) and a pattern (p
), implement regular expression matching with support for '.'
and '*'
where:
'.'
Matches any single character.'*'
Matches zero or more of the preceding element.The matching should cover the entire input string (not partial).
Example 1:
Input: s = "aa", p = "a" Output: false Explanation: "a" does not match the entire string "aa".
Example 2:
Input: s = "aa", p = "a*" Output: true Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
Example 3:
Input: s = "ab", p = ".*" Output: true Explanation: ".*" means "zero or more (*) of any character (.)".
Example 4:
Input: s = "aab", p = "c*a*b" Output: true Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches "aab".
Example 5:
Input: s = "mississippi", p = "mis*is*p*." Output: false
Constraints:
0 <= s.length <= 20
0 <= p.length <= 30
s
contains only lowercase English letters.p
contains only lowercase English letters, '.'
, and '*'
.'*'
, there will be a previous valid character to match.This is one of the top questions we should be solving in Dynamic Programming. The basic idea is to break the bigger task into smaller tasks and keep track of already solved tasks and re-use the result.
Time Complexity: O(m*n) // where m and n are the lengths of string and patterns
Space Complexity: O(m*n)